i need a box with a square base and an open top that cna hold 500 in^3. what is the least amount of material i can use to build this box

M(x)  =  452,56 in

Step-by-step explanation:

The volume of the open box is  500 in³

V = Area of the base times height

V(b) =  x² * h               where x is the side of the square and h the heigh

Then     500 =  x²*h

Total material to use is: material of the base + material of 4 sides

material of the base is  x²

material of one side is x*h  we have 4 sides then  4*x*h

Total material M(b)

M(b)  =  x²  +  4*x*h

And    as    h  =  500/ x²

M(x)  =  x²  +  4* x* 500/x²

M(x)  =  x²  +  2000/x

Tacking derivatives on both sides of the equation

M´(x)  =  2*x   - 200/x²

M´(x)  =  0         2*x  -  200/x²  = 0

x³ -  100  =  0

x³   =  100

x   =  4,64 in

And By substitution     h  =  500/x²

h  =  500/(4,64)²           h  =  23,22  in

How do we know that x = 4,64 make V(x) minimum

we get the second derivative

M´´(x)  =  2  +  200*2x/ x⁴    =  2  +  400/x³

M´´(x) is always positive

M´´(x) > 0     then M(x) has a minimum for  x= 4,64 in

The least amount of material is:

M(x)  =  x²    +    2000/x

M(x)  =  (4,64)²  *  2000/4,64

M(x)  = 21,53  +  431,03

M(x)  =  452,56 in

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons